Question

evaluate the limit, if it exists. if a limit does not exist, type \dne\.
\\(lim_{x
ightarrow - 6}\frac{\frac{1}{6}+\frac{1}{x}}{6 + x}\\)
limit

Explanation:

Step1: Combine fractions in numerator

First, combine $\frac{1}{6}+\frac{1}{x}$ using a common - denominator. The common denominator of 6 and $x$ is $6x$. So, $\frac{1}{6}+\frac{1}{x}=\frac{x + 6}{6x}$.
The limit becomes $\lim_{x
ightarrow - 6}\frac{\frac{x + 6}{6x}}{6 + x}$.

Step2: Simplify the complex - fraction

We can rewrite $\frac{\frac{x + 6}{6x}}{6 + x}$ as $\frac{x + 6}{6x(6 + x)}$. Since $x
eq - 6$ (we are taking the limit as $x$ approaches - 6, not setting $x=-6$), we can cancel out the non - zero factor $(x + 6)$ in the numerator and the denominator.
$\frac{x + 6}{6x(6 + x)}=\frac{1}{6x}$.

Step3: Evaluate the limit

Now, we find $\lim_{x
ightarrow - 6}\frac{1}{6x}$. Substitute $x=-6$ into $\frac{1}{6x}$.
$\lim_{x
ightarrow - 6}\frac{1}{6x}=\frac{1}{6\times(-6)}=-\frac{1}{36}$.

Answer:

$-\frac{1}{36}$